Question

If 400 lb of 40% cream are mixed with 800 lb of milk, and if 300 lb of the mixture is mixed with 300 lb of 0% skim, and the final mixture tests 8% fat, what does the 800 lb of milk test?

Answer 1

Sot: 40016 of 40% cream – cream content - 40 x 400 = 160 16 milk Content = 60 x 400 - 240 lb Let the cream 7. in 850 eb of milk be a'. cream content - x x 800 - 8x eb. milk content = 100-xx800 = (800 - 8x) ab Therefore, Total mixture contain - (160 + 8x) eb of cream and (240 + 800-8x) es of milk. with total = 800 + 400 = 1200. if 1200 eb of milk mixture contains (160+8 x) eb of cream then, 3 00 lb of milk mixture contains 160 +8x 200 of of of wean 1200 = (40+2x) eb of cream Cimilarly, if 1200 eb then 300 eb of of milk milk mixture mixture - contains (1040-8x) 16 of milk contains -&* x 3 co ub of milk R = (260-2x) ub of milk (Note: one can ie also find by from 300 just subtracting from the Total milk mir

when 300 lb of of skim & Added - cream content will be same as before in = (40+ 2x) eb Whereas, a milk content = 300 + (260-2x) eb. Total mixture = 300+ 300 = 600 eb. 40+25 x 100 600 = 8 or, 40+ 2n = 48 x= 4. Therefore, 800 eb of milk test had 4% of cream,